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alisha [4.7K]
3 years ago
11

What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures

?
Physics
1 answer:
Montano1993 [528]3 years ago
8 0

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.


Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2


If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?


In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.


Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2



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You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j
dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

            = mg=20\times 9.8 =196\ N

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0
3 years ago
When determining whether a chemical reaction has taken place, you observe and look for several indicators. Which would be consid
Inga [223]
The answer is B.

A cannot be the answer is melting is a physical change. No chemical reaction took place.

B is the answer as it is a EXOTHERMIC REACTION so heat will be given off.

C cannot be the answer as dissolving is basically atoms becoming ions, not a chemical reaction whereby a reactant reacts with another reactant to form a product.

D cannot be the answer. Same reason as for why A is not the answer.

Cheers.
6 0
3 years ago
Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would
HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js  is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle  ------------Quantum

Explanation:

 Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

  = 6.62607*10^-9/4.7*10^-26

  = 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

  =  6.62607*10^-9/4.44*10^-23

  = 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size =  6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  1*10⁻⁶* 1.1)

  =  6.62607*10^-9/1.1*10^-6

  = 6.02*10^-3 m

Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size =  22. cm = 0.22 m

V =  2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  0.71* 0.028)

  = 6.62607*10^-9/0.01988

   = 3.33*10^-7 m

Quantum Approach: The wavelength and the size are comparable

8 0
3 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
3 years ago
Why is argon gas used instead of air in light bulbs that contain a filament that is heated to glowing?
tamaranim1 [39]
<span>the heated filament will react with the oxygen in the air but now with the argon, which is a noble gas and hardly ever reacts.</span>
3 0
3 years ago
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